- #1
aihaike
- 55
- 0
Hi guys,
Einstein potential is define as
[tex]U_{e}=\sum_{i}\alpha_{i}(r_{i}-r0_{i})[/tex].
The partition function of the Hamiltonian
[tex]H=U_{e}+\frac{p^{2}}{2m}[/tex]
is given by
[tex]Q=\left(\frac{2m_{i}}{\alpha_{i}}\left(\frac{\pi}{\beta h}\right)^{2}\right)^{\frac{3}{2}}[/tex]
Which gives rises to the free energy
[tex]A=-\frac{3}{2\beta}\sum_{i=1}^{N}\ln\left[\frac{2m_{i}}{\alpha_{i}}\left(\frac{\pi}{\beta h}\right)^{2}\right][/tex]
Ok, now suppose we have two species in the system.
We can reformulate the partition function introducing [tex]\omega_{i}[/tex] and [tex]T_{\mathrm{E}_{i}}[/tex] for each species difine as
[tex]\omega_{i}=\sqrt{\frac{2\alpha_{i}}{m_{i}}}\quad\mathrm{and}\quad T_{\mathrm{E}_{i}}=\frac{h\omega_{i}}{2\pi k}[/tex]
And it comes
[tex]A=3N_{1}kT\ln\left(\frac{T_{\mathrm{E}_{1}}}{T}\right)+3N_{2}kT\ln\left(\frac{T_{\mathrm{E}_{2}}}{T}\right)[/tex]
where [tex]N_{1}[/tex] and [tex]N_{2}[/tex] are the number of atom of each species.
We also can take the quantum version of the partition function defines as
[tex]Q={\left(\frac{\exp\left(-\frac{T_{\mathrm{E}_{1}}}{2T}\right)}{1-\exp\left(-\frac{T_{\mathrm{E_{1}}}}{T}\right)}\right)^{3N_{1}}} {\left(\frac{\exp\left(-\frac{T_{\mathrm{E}_{2}}}{2T}\right)}{1-\exp\left(-\frac{T_{\mathrm{E_{2}}}}{T}\right)}\right)^{3N_{2}}}
[/tex]
Now comes my question.
On one hand it seems to me that the Einstein temperature [tex]T_{\mathrm{E}[/tex] is a characteristic is the system but it seems also depends on the system temperature.
I'm working on silica (SiO2) and I use to compute for each species [tex]\alpha[/tex] from the mean square displacement (u) calculated from a NVT monte carlo simulation (with the initial structure from the average coordinates of NPT simulation) using a standard 2-body potential with the formula
[tex]\alpha=\frac{3KT}{2u}[/tex]
The I calculate the corresponding frequency and Einstein temperature
[tex]\omega=\sqrt{\frac{2\alpha}{m}}\quad\mathrm{and}\quad T_{\mathrm{E}}=\frac{h\omega}{2\pi k}[/tex]
And the [tex]T_{\mathrm{E}[/tex] I get does not correspond to my apply temperature.
May be I should call it vibrational temperature instead of Einstein temperature.
I'm very confused, if we want to calculate the free energy of a crystal based on this model, what value of [tex]T_{\mathrm{E}[/tex] shoud we take? Is there table somewhere?
I also wonder if [tex]T_{\mathrm{E}[/tex] and [tex]\alpha[/tex] are really related actually.
Any comments are welcome.
Thanks in advance,
Eric.
Einstein potential is define as
[tex]U_{e}=\sum_{i}\alpha_{i}(r_{i}-r0_{i})[/tex].
The partition function of the Hamiltonian
[tex]H=U_{e}+\frac{p^{2}}{2m}[/tex]
is given by
[tex]Q=\left(\frac{2m_{i}}{\alpha_{i}}\left(\frac{\pi}{\beta h}\right)^{2}\right)^{\frac{3}{2}}[/tex]
Which gives rises to the free energy
[tex]A=-\frac{3}{2\beta}\sum_{i=1}^{N}\ln\left[\frac{2m_{i}}{\alpha_{i}}\left(\frac{\pi}{\beta h}\right)^{2}\right][/tex]
Ok, now suppose we have two species in the system.
We can reformulate the partition function introducing [tex]\omega_{i}[/tex] and [tex]T_{\mathrm{E}_{i}}[/tex] for each species difine as
[tex]\omega_{i}=\sqrt{\frac{2\alpha_{i}}{m_{i}}}\quad\mathrm{and}\quad T_{\mathrm{E}_{i}}=\frac{h\omega_{i}}{2\pi k}[/tex]
And it comes
[tex]A=3N_{1}kT\ln\left(\frac{T_{\mathrm{E}_{1}}}{T}\right)+3N_{2}kT\ln\left(\frac{T_{\mathrm{E}_{2}}}{T}\right)[/tex]
where [tex]N_{1}[/tex] and [tex]N_{2}[/tex] are the number of atom of each species.
We also can take the quantum version of the partition function defines as
[tex]Q={\left(\frac{\exp\left(-\frac{T_{\mathrm{E}_{1}}}{2T}\right)}{1-\exp\left(-\frac{T_{\mathrm{E_{1}}}}{T}\right)}\right)^{3N_{1}}} {\left(\frac{\exp\left(-\frac{T_{\mathrm{E}_{2}}}{2T}\right)}{1-\exp\left(-\frac{T_{\mathrm{E_{2}}}}{T}\right)}\right)^{3N_{2}}}
[/tex]
Now comes my question.
On one hand it seems to me that the Einstein temperature [tex]T_{\mathrm{E}[/tex] is a characteristic is the system but it seems also depends on the system temperature.
I'm working on silica (SiO2) and I use to compute for each species [tex]\alpha[/tex] from the mean square displacement (u) calculated from a NVT monte carlo simulation (with the initial structure from the average coordinates of NPT simulation) using a standard 2-body potential with the formula
[tex]\alpha=\frac{3KT}{2u}[/tex]
The I calculate the corresponding frequency and Einstein temperature
[tex]\omega=\sqrt{\frac{2\alpha}{m}}\quad\mathrm{and}\quad T_{\mathrm{E}}=\frac{h\omega}{2\pi k}[/tex]
And the [tex]T_{\mathrm{E}[/tex] I get does not correspond to my apply temperature.
May be I should call it vibrational temperature instead of Einstein temperature.
I'm very confused, if we want to calculate the free energy of a crystal based on this model, what value of [tex]T_{\mathrm{E}[/tex] shoud we take? Is there table somewhere?
I also wonder if [tex]T_{\mathrm{E}[/tex] and [tex]\alpha[/tex] are really related actually.
Any comments are welcome.
Thanks in advance,
Eric.